RevisedSOLUTIONSHW#4

RevisedSOLUTIONSHW#4 - SOLUTIONS (revision on No. 4)...

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SOLUTIONS (revision on No. 4) HOMEWORK NO. 4 CE 3400 – Introduction of Environmental Engineering Material Covered: Ch 7 (Energy Flows & Balances) and Ch 8 (Ecosystems) in Vesilind et al., 2010 Due Date: Monday, 11 October 2010 at 1:10 PM 1. This is Problem 7.3 in your textbook. A nuclear power station with an expected life of 25 years produces 750 MW/yr as useful energy. The energy cost is as follows: 150 MW lost in transmission 20 MW needed to mine the fuel 50 MW needed to enrich the fuel 80 MW (spread over 25 yr) to build the plant 290 MW lost as heat What is the efficiency of the plant itself? What is the efficiency of the entire generation system (including distribution losses and other investments)? Efficiency of plant = [ Useful energy OUT ] [Energy IN] × 100 Energy OUT = 750 MW/yr Energy IN = 750 MW/yr (useful energy) + 150 MW/yr (distribution losses) + 290 MW/yr (heat loss) Efficiency = 750 MW/yr 1190 MW/yr × 100 = 63 % I could also agree that this calculation only include heat loss. In that case, the efficiency is 72% Efficiency of system = [Useful energy OUT] [Total energy IN] × 100 Total energy IN ( in MW ) = 750 ( useful energy ) + 150 ( distribution losses ) + 20 ( mining ) + 50 ( enriching ) + 80 / 25 ( construction ) + 290 ( heat loss ) Efficiency of system = 750 MW 1263.2 MW × 100 = 59 % 2. This is Problem 8.2 in your textbook. A stream has a dissolved oxygen level of 9 mg/L, an ultimate oxygen demand of 12 mg/L, and an average flow of 0.2 m 3 /sec. An industrial waste with zero D.O., a BOD u of 20,000 mg/L and a flow rate of 0.006 m 3 /sec is discharged into the stream. What are the L o and D.O. in the stream immediately below the discharge point? Find: L o (BOD) and dissolved oxygen concentration in stream just below discharge point Assume steady state so: Q d = Q w + Q s = 0.2 m 3 /sec + 0.006 m 3 /sec = 0.206 m 3 /sec
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To calculate L o downstream, perform a mass balance on BOD values and solve for L o : L os Q os + L ow Q ow = L od Q od Rearranging and solving for L od : L od = L os Q os + L ow Q ow Q od = 12 mg / L (0.2 m 3 /sec) + 20,000 mg / L (0.0006 m 3 /sec) 0.206 m 3 /sec = 594 mg / L To calculate DO downstream, perform a mass balance on DO and solve for DO d : DO s Q s + DO w Q w = DO d Q d Rearranging and solving for DO d : DO d = DO s Q s + DO w Q w Q d = 9 mg / L (0.2 m 3 /sec) + 0 mg / L (0.0006 m 3 /sec) 0.206 m 3 /sec = 8.7 mg / L 3. This is Problem 8.7 in your textbook.
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This note was uploaded on 12/08/2010 for the course CE 2400 taught by Professor Colberg during the Spring '10 term at Wyoming.

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RevisedSOLUTIONSHW#4 - SOLUTIONS (revision on No. 4)...

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